12+1−3=

# $\frac{1}{\sqrt{2}+\sqrt{1}-\sqrt{3}}=$

1. A

$\frac{\left(1+\sqrt{3}-\sqrt{2}\right)\left(2+\sqrt{6}\right)}{4}$

2. B

$\frac{1+\sqrt{3}-\sqrt{2}}{-8}$

3. C

$\frac{2+\sqrt{6}}{4}$

4. D

none of these

Register to Get Free Mock Test and Study Material

+91

Live ClassesRecorded ClassesTest SeriesSelf Learning

Verify OTP Code (required)

### Solution:

$\begin{array}{l}\frac{1}{\left(\sqrt{2}-\sqrt{3}\right)+1}=\frac{1\left[\left(\sqrt{2}-\sqrt{3}\right)-1\right]}{{\left(\sqrt{2}-\sqrt{3}\right)}^{2}-{\left(1\right)}^{2}}=\frac{\sqrt{2}-\sqrt{3}-1}{2+3-2\sqrt{6}-1}=\frac{\sqrt{2}-\sqrt{3}-1}{4-2\sqrt{6}}\\ ⇒\frac{\left(\sqrt{2}-\sqrt{3}-1\right)\left(4+2\sqrt{6}\right)}{{\left(4\right)}^{2}-{\left(2\sqrt{6}\right)}^{2}}=\frac{\left(\sqrt{2}-\sqrt{3}-1\right)\left(4+2\sqrt{6}\right)}{16-24}\\ ⇒\frac{-2\left(1+\sqrt{3}-\sqrt{2}\right)\left(2+\sqrt{6}\right)}{-8}=\frac{\left(1+\sqrt{3}-\sqrt{2}\right)\left(2+\sqrt{6}\right)}{4}\end{array}$

## Related content

 Distance Formula Perimeter of Rectangle Area of Square Area of Isosceles Triangle Pythagoras Theorem Triangle Formulae Volume of Cylinder Perimeter of Triangle Formula Area Formulae Volume Formulae  